1. **State the problem:** Given the cubic equation $$x^3 - 9x^2 + 23x - 15 = 0$$ with roots $$\alpha, \beta, \gamma$$, find the value of $$\frac{1}{\alpha+2} + \frac{1}{\beta+2} + \frac{1}{\gamma+2}$$. 2. **Use the substitution:** Let $$y = x + 2$$, so $$x = y - 2$$. The roots $$\alpha, \beta, \gamma$$ correspond to $$y = \alpha + 2, \beta + 2, \gamma + 2$$. 3. **Rewrite the equation in terms of $$y$$:** Substitute $$x = y - 2$$ into the original equation: $$ (y - 2)^3 - 9(y - 2)^2 + 23(y - 2) - 15 = 0 $$ 4. **Expand and simplify:** $$ (y - 2)^3 = y^3 - 6y^2 + 12y - 8 $$ $$ (y - 2)^2 = y^2 - 4y + 4 $$ So, $$ y^3 - 6y^2 + 12y - 8 - 9(y^2 - 4y + 4) + 23(y - 2) - 15 = 0 $$ 5. **Distribute and combine like terms:** $$ y^3 - 6y^2 + 12y - 8 - 9y^2 + 36y - 36 + 23y - 46 - 15 = 0 $$ $$ y^3 - 15y^2 + (12y + 36y + 23y) + (-8 - 36 - 46 - 15) = 0 $$ $$ y^3 - 15y^2 + 71y - 105 = 0 $$ 6. **Identify the new roots:** The roots of this cubic are $$y_1 = \alpha + 2$$, $$y_2 = \beta + 2$$, $$y_3 = \gamma + 2$$. 7. **Sum of roots:** $$S = y_1 + y_2 + y_3 = 15$$ (coefficient of $$y^2$$ is $$-15$$, so sum is $$15$$). 8. **Sum of products of roots taken two at a time:** $$P = y_1 y_2 + y_2 y_3 + y_3 y_1 = 71$$. 9. **Product of roots:** $$Q = y_1 y_2 y_3 = 105$$ (constant term is $$-105$$, so product is $$105$$). 10. **Use the identity for sum of reciprocals:** $$ \frac{1}{y_1} + \frac{1}{y_2} + \frac{1}{y_3} = \frac{y_1 y_2 + y_2 y_3 + y_3 y_1}{y_1 y_2 y_3} = \frac{P}{Q} = \frac{71}{105} $$ **Final answer:** $$\boxed{\frac{71}{105}}$$